Question number 30. A barcode is composed of alternating black and white stripes, with the first and last stripes always black. Each stripe can be either 1 or 2 units wide, and the entire barcode has a width of 12. So here we have an example where we have some thin and thick strips. The first is black, it's thin. The last is black, it's thick. So the last one has width 2, the first one has width 1. So we have 6 thin strips and 3 thick strips, so the total width is 12. But 12 here is written as 6 plus 6, which the first 6 accounts for 6 times the strips of length 1, and the other 6 is 3 times the strips of length 2, for a total of 9 strips. And that's important to notice, that since the colors are alternating, that implies that there is always an odd number of strips, which we can illustrate pretty easily. So drawing an example barcode here, I can start with 2 of the end strips in black, and so there is here 1 strip to begin with, 1 strip to end with, and plus 1 here, strip for the white in the middle, even though it's thicker than 2. And we have an odd number of 3. Adding in a strip in the middle, let's do a black strip here, a thick one, in the middle, like that. And so what I have done now, is I have created plus 1 here for the number of black strips, and also plus 1 for the number of white strips, for a total of 5 now. So each time I draw in a black strip, I in fact create 2. Starting with the initial amount of 3, we just go up by 2 every time, so it's always an odd number. So always odd. And so now we can ask the question, how do we partition 12 into an odd number of 1's and 2's, because this is the only possibility. So we can start by saying 12 is equal to 10 plus 2, where here we have 10 strips of length 1, and 1 strip of length 2, and so that's a total of 11 strips. And the 2 here can be placed in any one of those 11 spots, so we have 11 choose 1, or 11 strips, like that. And then let's keep going and try to partition 12 into not 11 but 9 strips, and we can do that exactly by following our example here that's given in the problem. So let me just copy that. We have 6 plus 6, so that would be 6 times the 1's plus 3 times the 2's, and here we have a total of 9 strips, and that would be accounting for 9 choose 3, because the 3 2's can be placed in any of the 9 positions, which is 84 such strip arrangements, or 84 such barcodes. And finally, the last way to partition 12 into an odd number of strips is into 7 strips, and that would be 2 plus 10, but now we go the other way, 2 1's and 5 thick strips. So that's a total of 7 strips, and here that would account for 7 choose 5, which we compute to be 21 barcodes. And it is not possible to partition 12 into 5 strips, only of length 1 and 2, because, for example, 5 times 2 is 10, so already we're missing some width, and then we just add all of these up. So the total number here would be the strips of length 11, so that's one of them, the strips of length 9, that's 84 of them, and then the strips of length 7, that's 21 of them, for a total number of 116 possible barcodes, and so the answer here is 8, 116.

barcode calculation

stripe combinations

alternating stripes

width partitioning

unique arrangements